Lesson video

Hello, I'm Miss Miah and I'm so excited to be a part of your learning journey today.

I hope you enjoy this lesson as much as I do.

Today you will be able to multiply a three-digit by a one-digit number using expanded and short multiplication, this time involving multiple regrouping.

These are your keywords.

Short multiplication, expanded multiplication, regroup and/or regrouping.

Repeat them after me.

Short multiplication.

Expanded multiplication.

Regroup or regrouping.

Expanded multiplication is a way of recording the steps of calculation, focusing on partitioning one or more factors and showing partial products.

A way of recording using columns to set out and calculate a multiplication is called short multiplication.

The process of unitizing and exchanging between values is known as regrouping, for example, 10 ones can be regrouped for one 10.

One 10 can be regrouped for 10 ones.

So for this first lesson cycle, you'll be using expanded multiplication to explore this.

In this lesson, you'll meet Andeep and Izzy.

Now in the past you may have seen this, this is an example of expanded multiplication.

Here we can see that we're multiplying a two-digit by a one-digit number.

You may have also seen this, it's another example of expanded multiplication.

So today you will be exploring multiplying a three-digit by a one-digit number, including multiple instances of regrouping.

Now don't worry, it's not as complicated as it seems, we just need to break down the steps so you really understand how to tackle the questions, which involve multiple regrouping.

So one ticket to a rugby match costs £148, Eric buys six, what is the total cost? What is the multiplication equation needed? Have a think.

Well, if you've got 148 multiplied by six or six multiplied by 148, you are correct.

Now, we always place the largest factor at the top and we place the smallest factor at the bottom and we're going to begin by multiplying the ones.

So six multiplied by eight is 48 ones, now the 48 can be regrouped as four 10s and eight ones.

What we do is we place the four in the 10s column and eight in the ones column.

What do you notice? We have regrouped our ones into our 10s.

Now we're going to move on to our 10s.

Six multiplied by four 10s is equal to 24 10s.

Now 24 10s can be regrouped as 200, four 10s, and zero ones.

And when it comes to writing this in our method, this is what it looks like.

So we place the two in the hundreds column, four in the 10s column and zero as your placeholder.

What do you notice this time? Again we have regrouped, but this time in our 10s to our hundreds.

Now we're going to move on to the hundreds column.

So six multiplied by 100 is equal to six 100s.

We write the six in the hundreds column and then our zero as our placeholder in the 10s and ones column.

Now we combine our partial products, I'm sure you knew that.

So what we do, it's it's very similar to column edition and the product that you should have got is 888.

So six tickets would cost £888.

I really like the expanded multiplication method, especially when it comes to recording our regrouping, because it's very easy to lay it out and see where the regrouping has happened.

Adding the partial products afterwards is just one further step that we have to do as compared to short multiplication, but again, it's very useful to use expanded multiplication if you haven't got the hang of short multiplication yet.

Over to you, I'd like you to fill in the gaps.

You can pause the video here.

So how did you do? If you got the digits on the screen correct, give yourself a tick.

You have correctly identified the partial products for this question.

Let's move on now.

One laptop costs £743.

Eric has £2,180, will he be able to purchase three? Now, what is the multiplication equation needed? So I want you to think about what is known, what is unknown, that may help you to think about the equation required.

So the equation we're calculating is 743 multiplied by three, because there are three laptops.

We're going to start by multiplying our ones, three multiplied by three ones is nine ones.

So we can place the nine in the ones column.

Then we're going to move on to our 10s column.

Three, multiplied by four 10s is? If you've got 12 10s, you are correct.

So 12 10s can be regrouped as 102 10s.

How do you think we're going to record this? Well, we'll place the one in the hundreds column, a two in the 10s column and zero as our placeholder.

If you knew that, good job.

Now we must multiply our hundreds.

What do you notice? We have regrouped from our 10s to our hundreds, it's clearly recorded, we've got 102 10s and zero ones.

Now we can move on to multiplying in our hundreds column.

Three multiplied by seven equals 21 hundreds.

21 hundreds can be regrouped as 2,100.

How do you think we're going to record this? Well, what we have to do is place the two in the thousands column, the one in the hundreds and zero as your placeholder in the 10s and the ones.

So there you go, that's what it looks like.

What do you think we have to do next? Let's have a look.

Yes, we have to combine our partial products.

So we've got three partial products here.

This part is similar to column edition, so 2,229 is our product.

The total cost of three laptops is £2,229.

So as Eric has £2,180, he will not be able to purchase all three laptops, he'll only be able to purchase two.

Over to you.

I want you to have a look at this expanded multiplication and I want you to think, where did you regroup? You could pause the video here.

How did you do? By looking at our expanded multiplication, we can see that we regroup in the 10s and hundreds column.

Izzy is solving the calculation below, "I think I will regroup in the ones column only." Do you agree? And also I'd like you to explain your thinking to your partner.

Let's calculate this together.

Four multiplied by five gives us 20 ones.

20 ones can be regrouped as two 10s, so we've already had one regroup in the ones column.

Four multiplied by two 10s gives us eight 10s, there's no regrouping required in the 10s column, however, four multiplied by five 100s is 20 100s.

Regrouping is also required in the hundreds column as 20 100s is greater than ten 100s, so we must regroup and this can be written as 2000.

What do you notice? Regrouping in the hundreds has resulted in a four digit partial product.

Sometimes there are surprise regrouping, like in this example, there are two instances.

Over to you which multiplication will need regrouping in the hundreds and 10s column, is it A, B, or C? And I'd like you to explain your thinking to your partner.

You can pause the video here.

So how did you do? If you got A, you are correct, because three multiplied by one is three ones, three multiplied by five 10s is 15 10s, so we must regroup and then three multiplied by four 100s is 12 100s.

Again, it requires regrouping, back to you.

This time I'd like you to match the equation to the correct partial product expression.

You've got four equations here and you've got four partial product expressions on the right.

You can pause the video here, off you go.

So how did you do? If you got this as your answer, you are correct.

Give yourself a tick.

Let's move on.

So onto your tasks for the first learning cycle.

For this first task, you are going to take the correct instances of regrouping for each multiplication equation.

And you've got five equations that you will be doing this for.

And for task two, you are going to complete the word problem using expanded multiplication.

So the question is, Andeep, Jacob and Izzy are given bags of marbles.

Find the total amount of marbles for each person who has the least? So Jacob has got eight bags and in each bag there's 345 marbles.

Andeep has six bags and in each bag there are 453 marbles.

And lastly, Izzy has five bags and in each bag there are 403 marbles.

Think about the multiplication equations required to help you solve this problem.

You can pause the video here, good luck.

How did you do? If you managed to tick all the correct boxes, well done.

I'm going to leave this on the screen for a bit so you can check.

For task two, if you got this calculation as your answer, well done, you've got it correct.

So Jacob would've had 2,760 marbles altogether.

Andeep would've had 2,718 marbles altogether and lastly, Izzy would've had 2,015 marbles altogether, which means that Izzy has the least amount of marbles.

Let's move on.

For this lesson cycle, you are going to be using short multiplication.

One ticket costs, £148, Eric buys six.

What is the total cost and what is the multiplication equation needed? So first you're going to place the largest factor at the top, which is 148.

You're then going to place the smallest factor at the bottom, which is six.

Six multiplied by eight ones is 48 ones.

So firstly, 48 ones can be regrouped as four 10s and eight ones, so it's very important we regroup.

So with this four 10s, we must remember to place it underneath the 10s column and then replace the remaining eight ones in the ones column.

Next we move on to the 10s column.

So six multiplied by four 10s is 24 10s.

24 10s add the regrouped four 10s is 28 10s.

28 10s can be regrouped as two 100s and eight 10s.

So we place the two in the hundreds column, but we must remember to place it underneath and we place the eight 10s, that's it, you've got it, in the 10s column.

Now we move on to the hundreds column, fantastic.

So we must multiply six by 100, which is six 100s.

What do I need to do now? Have a think.

That's it, so we need to add the remaining two 100s.

So six, add the regrouped two 100s is 800.

So we place the eight in the hundreds column.

Six tickets would cost £888.

Over to you, I'd like you to fill in the gaps.

You can pause the video here.

How did you do? So if you got this as your answer, you are correct, well done.

One laptop costs £743, Eric has £3,700, will he be able to purchase five laptops? Want you to think about what is the multiplication equation needed? So this is Andeep's calculation and this is Izzy's calculation.

Who is correct and how do you know? Well, let's have a look.

Andeep is incorrect, because he did not regroup the 10s into the hundreds correctly.

The total cost is actually £3,715.

Over to you, which calculation will require three instances of regrouping? Is it A, B, or C? You can pause the video here to have a think.

So what did you get? If you got C as your answer, you are correct and that's because when we multiply each digit by five, we are ending up more than our base 10 value, so we are going to have to regroup.

Izzy used the oven 525 times last year.

Each time she used a baking tray that allowed her to bake four cupcakes at a time.

Izzy now wants to calculate how many cupcakes she baked altogether.

And you can see that there's a short multiplication recording here.

Is Izzy correct? I'd like you to explain your thinking to your partner.

Let's calculate together what mistake might have Izzy made? I want you to think about that as we are doing our calculation together.

So let's start off with multiplying our ones.

We end up with 20 ones, we must regroup the two 10s and place it underneath the 10s column and we can see that Izzy's done that as well.

We look at our 10s and we multiply within our 10s.

So four multiplied by two 10s is eight 10s.

But what do we do afterwards? We must add the regrouped two 10s.

So eight 10s add two 10s is 10 10s.

And then we must regroup and place 100 in our hundreds column and zero in our 10s column.

So here we've got zero in our 10s column and 100 in our hundreds column, but it's placed underneath because we regrouped.

So what Izzy actually forgot to do was add the regrouped amount, which was two 10s and that's a very common mistake, I made it all the time when I was younger, make sure you add the regrouped amount.

And then if we carry on, four multiplied by 500 is 20 100s.

This can be regrouped as 2000, so we put two in our thousands column and then we must add the regrouped 100 and we can place that in the hundreds column.

So Izzy baked 2,100 cupcakes altogether.

Onto your final tasks for lesson cycle two.

So for this part, you are going to be using short multiplication to record and calculate these equations.

A is 356 multiplied by eight, B is 536 multiplied by six and C is seven multiplied by 503.

Think very carefully about how you arrange your numbers and the alignment of your digits when regrouping.

And task two involves problem solving, my favourite, so Jacob, Andeep and Izzy took part in a baking competition.

Whoever baked the most won.

They each used different baking trays.

So what you are going to do is find the total amount of cupcakes for each person.

You are also going to tell me who baked the least and you're going to make sure you use short multiplication.

So we can see here that Jacob baked eight at a time and he used the oven 346 times.

Andeep baked six at a time and he used the oven 456 times.

And lastly, Izzy baked five at a time and she used the oven 406 times.

You can pause the video here, off you go, good luck.

So how did you do? For the first task, if you've got these as your answers, you are correct.

I'm going to just leave this on the screen for a bit so you can mark your work.

If you've got all three correct, really good job, well done.

Let's move on to the second task.

Now Jacob baked 2,768 cupcakes altogether, if you've got, that smashing job.

Andeep baked, 2,736 cupcakes all together.

So that's the calculation that you should have had.

If you managed to get that, good job.

And lastly, Izzy baked 2,436 cupcakes altogether and as a conclusion, she baked the least.

If you managed to get this multiplication equation correct, good job.

So we've reached the end of the lesson, well done for making it through.

I really enjoyed this lesson and I hope you did too.

Let's summarise, so today you multiplied a three digit by a one digit number with multiple regroups.

You can now hopefully multiply a three digit by a one digit number, including multiple regroups.

You can now also choose an efficient method to solve multiplication problems. I really love the short multiplication method.

What strategy do you prefer? Hope to see you in the next video.